Step-by-step explanation:
18.
the last one.
-4.9t² + 24.5t + 117.6 = 0
the ball starts at 117.6 meters height. then, during the first seconds the thrust upwards is adding height, until finally, with more and more seconds passing, gravity will win with a vengeance and pulls the ball down faster and faster than any other force in any other direction.
19.
the general solution for a quadratic equation
y = ax² + bx + c
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = t
a = -4.9
b = 24.5
c = 117.6
t = (-24.5 ± sqrt(600.25 - -4×4.9×117.6))/-9.8 =
= (-24.5 ± sqrt(600.25 + 2304.96))/-9.8 =
= (-24.5 ± sqrt(2905.21))/-9.8 = (-24.5 ± 53.9)/-9.8
t1 = (-24.5 + 53.9)/-9.8 = -3
t2 = (-24.5 - 53.9)/-9.8 = 8
a negative time does not make any sense, so the real solution is : the ball reached the ground after 8 seconds.
20.
now the equation is
-4.9t² + 24.5t + 117.6 = 49
but this is quickly transformed again into a "= 0" equation :
-4.9t² + 24.5t + 68.6 = 0
t = (-24.5 ± sqrt(600.25 - -4×4.9×68.6))/-9.8 =
= (-24.5 ± sqrt(600.25 + 1344.56))/-9.8 =
= (-24.5 ± sqrt(1944.81))/-9.8 = (-24.5 ± 44.1)/-9.8
t1 = (-24.5 + 44.1)/-9.8 = -2
t2 = (-24.5 - 44.1)/-9.8 = 7
again, negative time does not make sense.
the ball will be at 49 m height after 7 seconds.
